Estimating the shape parameter of a Pareto distribution under restrictions

Abstract

In this paper estimation of the shape parameter of a Pareto distribution is considered under the a priori assumption that it is bounded below by a known constant. The loss function is scale invariant squared error. A class of minimax estimators is presented when the scale parameter of the distribution is known. In consequence, it has been shown that the generalized Bayes estimator with respect to the uniform prior on the truncated parameter space dominates the minimum risk equivariant estimator. By making use of a sequence of proper priors, we also show that this estimator is admissible for estimating the lower bounded shape parameter. A class of truncated linear estimators is studied as well. Some complete class results and a class of minimax estimators for the case of an unknown scale parameter are obtained. The corresponding generalized Bayes estimator is shown to be minimax in this case as well.

Appendix

Proof of Theorem 2.1

The risk difference the estimators $\delta _{g}(T)$ and $\delta _{BS}(T)$ is given by

$$\begin{aligned} \varDelta (\tau )= & {} R(\tau , \delta _{g}) - R(\tau , \delta _{BS})\\= & {} 2\int _{0}^{\infty }g'(y)\Big [\int _{y}^{\infty }\Big (\frac{\delta _{BS} (t)\,g(y)}{\tau }-1 \Big )\, \frac{\delta _{BS}(t)}{\tau } \, f(t)dt\Big ]dy. \end{aligned}$$

Since $g(y)$ is nondecreasing, $\varDelta (\tau )$ will be nonpositive if

$$\begin{aligned} g(y)\le & {} \frac{\tau }{n-2} \, \frac{\int _{y}^{\infty }\, t^{n-2}\, e^{-\tau t}\, dt}{\int _{y}^{\infty }\, t^{n-3}\, e^{-\tau t}\, dt},~~\text{ for } \text{ all }~y>0 ~\text{ and }~ \tau \ge \tau _0. \end{aligned}$$

In the last equality we use the transformation $\tau \,t = u$ to obtain

$$\begin{aligned} g(y)\le & {} \frac{1}{n-2}\, E_{\tau ^{*}} (U) \end{aligned}$$

where $\tau ^{*} = \tau \, y, ~ \tau _0^{*} = \tau _{0}\, y$. The expectation is taken with respect to the density $f^{*}(u) = u^{n-3}\, e^{- u}$ truncated on $(\tau ^{*}, \infty )$. Observe that $f^{*}(u)$ has increasing MLR in $\tau$. Thus we have

$$\begin{aligned} g(y)\le & {} \frac{1}{n-2}\,\inf _{\tau ^{*}} E_{\tau ^{*}} (U) = \frac{1}{n-2}\, E_{\tau _{0}^{*}} (U). \end{aligned}$$

So

$$\begin{aligned} g(y)\le & {} \frac{1}{n-2}\, \frac{\int _{\tau _{0}^{*}}^{\infty }\, u^{n-2}\, e^{- u}\, du}{\int _{\tau _{0}^{*}}^{\infty }\, u^{n-3}\, e^{- u}\, du} = g_{0}(y). \end{aligned}$$

This completes the proof of the theorem.

Consider the following functions

$$\begin{aligned} g_{m}(t) = e^{-\tau _0\, t}\, t^{m-1},~~~G_{m}(t) = \int _{t}^{\infty }\, u^{m-1}\, e^{-\tau _0\, u}\, du. \end{aligned}$$

Lemma 6.1

$\frac{g_{m}(t)}{\tau _0\, G_{m}(t)} \le 1$   for all $m\ge 1$.

Proof

For $m\ge 1$

$G_{m}(t) = \int _{t}^{\infty }\, u^{m-1}\, e^{-\tau _0\, u}\, du \ge t^{m-1}\, \int _{t}^{\infty }\,e^{-\tau _0\, u}\, du = \frac{1}{\tau _0}\, t^{m-1}\, e^{-\tau _0\, t} = \frac{g_{m}(t)}{\tau _0}$.

As a consequence of the above lemma we have the following result.

Lemma 6.2

$$\begin{aligned} E\Big (\frac{1}{\tau ^l T^p}\Big )\, \Big (\frac{g_{m}(t)}{G_{m}(t)}\Big )^k \le \frac{(\tau _0)^{k-l+p}}{1+(l-p)\xi }\, \frac{\varGamma (n-p)}{\varGamma (n)},~~n\ge p,~~l\ge p, \end{aligned}$$

and if $p=l$ then

$$\begin{aligned} E\Big (\frac{1}{\tau T}\Big )^l\, \Big (\frac{g_{m}(t)}{G_{m}(t)}\Big )^k \le (\tau _0)^k\, \frac{\varGamma (n-l)}{\varGamma (n)},~~n\ge l, \end{aligned}$$

where the expectation is taken with respect to the joint distribution of $X$ and $\tau$.

Proof

$$\begin{aligned} E\Big (\frac{1}{\tau ^l T^p}\Big )\, \Big (\frac{g_{m}(t)}{G_{m}(t)}\Big )^k \!= & {} \! \frac{\tau _0^{\xi ^{-1}}}{\xi } \int _{\tau _0}^{\infty } \tau ^{-1-\xi ^{-1}-l} \Big [\frac{\tau ^n}{\varGamma {(n)}}\int _{0}^{\infty } \Big (\frac{g_{m}(t)}{G_{m}(t)}\Big )^k e^{-\tau t} t^{n-p-1} dt\Big ] d\tau \\\le & {} \frac{\varGamma {(n-p)}}{\varGamma (n)}\,\frac{\tau _0^{k + \xi ^{-1}}}{\xi }\, \int _{\tau _0}^{\infty } \tau ^{-1-\xi ^{-1}-l+p} d\tau \nonumber \\= & {} \frac{(\tau _0)^{k-l+p}}{1+(l-p)\xi }\, \frac{\varGamma {(n-p)}}{\varGamma (n)}. \end{aligned}$$

Lemma 6.3

$E\Big (\frac{U(T)}{T}\Big ) = -\frac{1}{n-1} E(U'(T)) + \frac{\tau }{n-1}\, E(U(T))$, $T \sim G(n, \tau )$

Proof

$$\begin{aligned} E\Big (\frac{U(T)}{T}\Big )= & {} \frac{\tau ^n}{\varGamma (n)}\, \int _{0}^{\infty } u(t) t^{n-2} e^{-\tau t}\, dt = \frac{\tau ^n}{\varGamma (n)}\, \int _{0}^{\infty } u(t) e^{-\tau t} \frac{1}{n-1} \, d(t^{n-1}) \\= & {} \frac{\tau ^n}{\varGamma (n)} \Big [ \frac{1}{n-1} u(t) e^{- \tau t} t^{n-1} \Big |_{t=0}^{+ \infty } - \int _{0}^{\infty } \frac{1}{n-1} \Big (u'(t) e^{- \tau t} - u(t) \tau e^{- \tau t}\Big ) \Big ] \\= & {} -\frac{1}{n-1} E(U'(T)) \!+\! \frac{\tau }{n-1}\, E(U(T)), \hbox { considering that } \lim _{t ->0}u(t) e^{- \tau t} t^{n-1} \\= & {} \lim _{t -> +\infty } u(t) e^{- \tau t} t^{n-1} = 0. \end{aligned}$$

Lemma 6.4

$\lim _{\xi \rightarrow \infty } ({r_{\xi }}(\delta _{GB}) - r_{\xi }(\delta _{\xi }) )= 0.$

Proof

$$\begin{aligned} r_{\xi }(\delta _{\xi })= & {} \int _{\tau _0}^{\infty }\, \pi _{\xi }(\tau ) \int _{0}^{\infty }\,\Big (\frac{n_{\xi } - 2}{\tau t} - 1 + \frac{g_{n_{\xi } - 2}(t)}{\tau G_{n_{\xi } - 2}(t)}\Big )^2\, f(t, \tau ) dt d\tau \\= & {} \Big (\frac{(n_{\xi } - 2)^2}{(n-1)(n-2)} - 2 \frac{n_{\xi } - 2}{n-1} + 1 \Big ) \\&+ \int _{\tau _0}^{\infty }\, \pi _{\xi }(\tau ) \int _{0}^{\infty }\, \Big (\frac{g_{n_{\xi } - 2}(t)}{\tau G_{n_{\xi } - 2}(t)}\Big )^2 \, f(t, \tau ) dt d\tau \\&+ 2 \int _{\tau _0}^{\infty }\, \pi _{\xi }(\tau ) \int _{0}^{\infty }\, \Big (\frac{n_{\xi } - 2}{\tau t} - 1\Big ) \frac{g_{n_{\xi } - 2}(t)}{\tau G_{n_{\xi } - 2}(t)}\, f(t, \tau ) dt d\tau . \end{aligned}$$

In a similar way,

$$\begin{aligned} r_{\xi }(\delta _{GB})= & {} \int _{\tau _0}^{\infty }\, \pi _{\xi } (\tau ) \int _{0}^{\infty }\,\Big (\frac{n - 2}{\tau t} - 1 + \frac{g_{n - 2}(t)}{\tau G_{n - 2}(t)}\Big )^2\, f(t, \tau ) dt d\tau \\= & {} \frac{1}{n-1} + \int _{\tau _0}^{\infty }\, \pi _{\xi }(\tau ) \int _{0}^{\infty }\, \Big (\frac{g_{n - 2}(t)}{\tau G_{n - 2}(t)}\Big )^2 \, f(t, \tau ) dt d\tau \\&+ 2\int _{\tau _0}^{\infty }\, \pi _{\xi }(\tau ) \int _{0}^{\infty }\, \Big (\frac{n- 2}{\tau t} - 1\Big ) \frac{g_{n - 2}(t)}{\tau G_{n - 2}(t)}\, f(t, \tau ) dt d\tau . \end{aligned}$$

Considering that,

• 1. $\displaystyle n_{\xi } = n - \frac{1}{\xi } \rightarrow _{\xi \rightarrow + \infty } n$

• 2. $\displaystyle g_{n_{\xi }-2}(t) = e^{- \tau _0 t} t^{n_{\xi } - 2 - 1} \rightarrow _{\xi \rightarrow + \infty } e^{- \tau _0 t} t^{n - 2 - 1} = g_{n-2}(t)$

• 3. $\displaystyle G_{n_{\xi }-2}(t) \!=\! \int _t^{+ \infty } \, e^{- \tau _0 u} u^{n_{\xi } - 2 - 1} du \rightarrow _{\xi \rightarrow + \infty } \int _t^{+ \infty } \, e^{- \tau _0 u} u^{n - 2 - 1} du = G_{n-2}(t)$.

It is obvious that, $\lim _{\xi \rightarrow \infty } ({r_{\xi }}(\delta _{GB}) - r_{\xi }(\delta _{\xi }) ) = 0.$